# How do you solve 8/6=(k-4)/(k-3)?

Feb 22, 2017

See the entire solution process below:

#### Explanation:

Multiply each side of the equation by $\textcolor{red}{6} \left(\textcolor{b l u e}{k - 3}\right)$ to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{6} \left(\textcolor{b l u e}{k - 3}\right) \times \frac{8}{6} = \textcolor{red}{6} \left(\textcolor{b l u e}{k - 3}\right) \times \frac{k - 4}{k - 3}$

$\cancel{\textcolor{red}{6}} \left(\textcolor{b l u e}{k - 3}\right) \times \frac{8}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} = \textcolor{red}{6} \cancel{\left(\textcolor{b l u e}{k - 3}\right)} \times \frac{k - 4}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{k - 3}}}}$

$8 \left(\textcolor{b l u e}{k - 3}\right) = \textcolor{red}{6} \left(k - 4\right)$

Next, expand the terms within parenthesis on each side of the equation:

$\left(8 \times \textcolor{b l u e}{k}\right) - \left(8 \times \textcolor{b l u e}{3}\right) = \left(\textcolor{red}{6} \times k\right) - \left(\textcolor{red}{6} \times 4\right)$

$8 k - 24 = 6 k - 24$

Then, add $\textcolor{red}{24}$ and subtract $\textcolor{b l u e}{6 k}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$8 k - 24 + \textcolor{red}{24} - \textcolor{b l u e}{6 k} = 6 k - 24 + \textcolor{red}{24} - \textcolor{b l u e}{6 k}$

$8 k - \textcolor{b l u e}{6 k} - 24 + \textcolor{red}{24} = 6 k - \textcolor{b l u e}{6 k} - 24 + \textcolor{red}{24}$

$2 k - 0 = 0 - 0$

$2 k = 0$

Now, divide each side by $\textcolor{red}{2}$ to solve for $k$ while keeping the equation balanced:

$\frac{2 k}{\textcolor{red}{2}} = \frac{0}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} k}{\cancel{\textcolor{red}{2}}} = 0$

$k = 0$